Quadrilaterals Exercise 8.1
Maths Chapter 8 of Class 9 NCERT book is Quadrilaterals and the chapter seems quite tricky to the ninth graders especially in the beginning.
It is important that your last year concepts on polygon are clear.
What is a Quadrilateral?
You should known that a quadrilateral is a polygon, that is, a closed figure made up of four sides. You should also be aware of differences between quadrilaterals and parallelograms.
What is the angle sum property of a Quadrilateral?
NCERT solutions for class 9 Maths Chapter 8 Exercise 8.1
Let the angles of the Quadrilateral be 3x, 5x, 9x and 13x
Using Angle sum property of the quadrilateral
3x + 5x + 9x + 13x = 3600
30x = 3600
x= 120
So the 4 angles of the Quadrilateral are
3x = 3 x 120 = 360
5x = 5 x 120 = 600
9x = 9 x 120 = 1080
13x = 13 x 12 = 1560
Given: ABCD is a parallelogram with equal diagonals
To prove: The parallelogram is a rectangle
Proof:
In Triangle ABC and BCD
BC = BC (Common)
AC = BD (Diagonals are equal; given)
AB = CD (opposite sides of a parallelogram are equal)
By SSS, ABC≅ BCD
∠ABD = ∠BCD (by CPCT) …. (i)
But since, angles ∠ABD and ∠BCD are interior angles, they are supplementary
That is, ∠ABD + ∠BCD = 1800
2 ∠ABD = 1800
∠ABD = 900 = ∠BCD
Since in a parallelogram all opposite angles are equal, that means all angles measure 900
[A RECTANGLE is a parallelogram with equal diagonals and all its angles measure 900. Therefore it is a rectangle. Hence proved
Given: A quadrilateral ABCD whose diagonals bisect at 900
To prove: The quadrilateral is a rhombus
Proof: In Triangles AOB and BOC
AO =OC (Diagonals bisect each other)
OB = OB (Common)
∠AOB = ∠BOC (Given, both 900)
Triangle AOB ≅ Triangle BOC by SAS Congruency rule
AB = BC (by CPCT) ….. (1)
In Triangles BOC and COD
BO =OD (Diagonals bisect each other)
OC = OC (Common)
∠BOC = ∠COD (Given, both 900)
Triangle BOC≅ Triangle COD by SAS Congruency rule
BC = CD (by CPCT) …… (2)
Similarly triangle AOD ≅ Triangle DOC
AD = CD (by CPCT) ….. (3)
By (1), (2) and (3)
AB = BC = CD = AD and
Its diagonals bisect at 900
Therefore it’s a rhombus
Given: ABCD is a square
To prove: Its diagonals are equal and bisect each other
Proof: In Triangles ABC and BCD
AB = CD (all sides of a square are equal)
BC = BC (Common)
∠B = ∠D (Angles of a square are 900)
ABC ≅BCD by SAS criteria
AC = BD (by CPCT) … (1)
In Triangles AOB and COD
AB = CD (all sides of a square are equal)
∠ABO = ∠ODC (Alternate angles)
∠BAO = ∠OCD (Alternate angles)
Triangle AOB ≅Triangle COD by ASA criteria
OA = OC & OB = OD (by CPCT) … (2)
In Triangles AOB and BOC
OA =OC (Proved above)
OB = OB (Common)
AB = BC (all sides of square are equal)
Triangle AOB ≅Triangle BOC by SSS criteria
∠AOB = ∠BOC (by CPCT)
These angles form linear pair
∠AOB + ∠BOC = 1800
2∠AOB = 1800
∠AOB = 900 = BOC …. (3)
By (1), (2) and (3)
Diagonals of a square are equal and they bisect each other at 900
Given: A quadrilateral with equal diagonals that bisect each other at 900
To prove: It is a square
Proof:
In Triangles AOB and COD
OA =OC (GIVEN)
OB = OD (Given)
∠AOB = ∠COD (Corresponding angles)
Triangle AOB ≅Triangle COD by SAS criteria
AB = CD (by CPCT) … (1)
Similarly Triangle AOD ≅Triangle BOC by SAS criteria
AD = BC (by CPCT)…. (2)
In Triangles AOB and BOC
OA =OC (Given)
OB = OB (Common)
∠AOB = ∠BOC = 900 (Given; diagonals bisect at 900)
Triangle AOB ≅Triangle BOC by SAS criteria
AB= BC (by CPCT) …. (3)
By (1), (2) and (3) AB = BC = CD = AD
Therefore, the quadrilateral is a square.
Given: ABCD is a parallelogram and its diagonal AC bisects A i.e. BAC = DAC
To prove: (i) It bisects A also
(ii) ABCD is a rhombus
Proof: (i) DC//AB and AC is the transversal
∠BAC = ∠DCA (Alternate Angles) … (1)
Similarly AD//BC and AC is the transversal
∠DAC = ∠BCA (Alternate Angles) …. (2)
∠BAC = ∠DAC (given) … (3)
By (1), (2) and (3)
∠DCA = ∠BCA
Therefore AC also bisects C
(ii) Construct other diagonal
In triangles AOD and AOB
OA = OA (Common)
∠BAC = ∠DAC (given)
DO =OB (Diagonals of a //gm bisect each other)
Triangle AOD ≅ Triangle AOB by SSA criteria
AB = BC …. (3)
∠AOD = ∠AOB (by CPCT)
∠AOD & ∠AOB form linear pair
∠AOD + ∠AOB = 1800
2∠AOD = 1800
∠AOD = ∠AOB = 900 … (4)
By 3, the //gm’s adjacent sides are equal, that means all sides are equal, and by (4) its diagonals bisect at 900, therefore this //gm is a rhombus.
Given: ABCD is a rhombus
To prove: Diagonal AC bisects A as well as C and diagonal BD bisects B and D
Proof: In triangles AOB and BOC
AO =OC (Diagonals of a rhombus bisect each other)
∠ABO = ∠OBC (Equal sides have opposite angles equal) … (1)
Similarly in triangles AOD and DOC
AO =OC (Diagonals of a rhombus bisect each other)
∠ADO = ∠ODC (Equal sides have opposite angles equal) … (2)
By (1) and (2), BD bisects D as well as B
Similarly in triangles AOB and AOD
OB = OD (Diagonals of a rhombus bisect each other)
∠BAO = ∠OAD (Equal sides have opposite angles equal) … (3)
In triangles BOC and COD
OB = OD (Diagonals of a rhombus bisect each other)
∠BCO = ∠OCD (Equal sides have opposite angles equal) … (4)
By (3) and (4), AC bisects both C as well as A
Given: ABCD is a rectangle where diagonal AC bisects C as well as A i.e. ∠BAO = ∠OAD and ∠BCO = ∠OCD
To prove: (i) ABCD is a square
(ii) Diagonal BD bisects B and D
Proof:
In triangles AOB and AOD
OB = OD (diagonals of a rectangle bisect each other)
OA = OA (Common)
∠BAO = ∠OAD (given)
Triangle AOD ≅ Triangle AOB by SSA criteria
AB = AD (by CPCT) … (1)
But it is a rectangle where AB= CD and BC = AD
By (1), AB = BC = CA = DA … (2)
∠AOB = ∠AOD (by CPCT)
But ∠AOB & ∠AOD form linear pair
∠AOB + ∠AOD = 1800
2∠AOB = 1800
AOB = AOD = 900 … (2) (i.e. the diagonals bisect at 900)
By (1) and (2), it becomes clear that this rectangle is a square.
(ii) We know that OA = OC for triangles AOB and BOC as well as in triangles AOD and DOC
∠ABO = ∠OBC and ∠ADO = ∠DOC (Equal sides have opposite angles equal)
This proves that BD not only bisects AC but also angles B and D
Given: ABCD is a //gm and DP = BQ
To prove:
Proof:
In triangles APD and CQB
AD = BC (Opposite sides of //gm ABCD are equal)
DP = BQ (Given)
∠QBC = ∠ADP (AD//BC, BD is transversal; they are alternate angles)
Triangle APD ≅ Triangle CQB by SAS Criteria
AP = CQ (by CPCT)
In triangles AQB and CPD
AB = DC (Opposite sides of //gm ABCD)
DP = BQ (Given)
∠QBA = ∠CDP (AB//DC, BD is transversal; they are alternate angles)
Triangle AQB ≅ Triangle CPD by SAS Criteria
AQ= CP (by CPCT)
AP = CQ and AQ= CP (proved above)
A =C and Q=P (equal sides have opposite angles equal)
Therefore AQCP is a //gm
Given: ABCD is a //gm
To Prove:
Proof:
In triangles APB and CQD
AB = CD (Opposite sides of //gm ABCD are equal)
∠ABP = ∠BDC (alternate angles; AB//CD, BD is transversal)
∠APB = ∠DQC (Both 900)
Triangle APB ≅Triangle CQD by AAS Criteria
AP = CQ (by CPCT)
Given:
To prove:
Proof:
If AB = DE and AB //DE, then the distance between these line segments will also be equal (i.e. AD = BE) and //. Therefore ABED is a //gm
If BC = EF and BC //EF, then the distance between these line segments will also be equal (i.e. CF = BE) and //. Therefore BEFC is a //gm
By (i) and (ii) AD//BE and BE//CF, Also AD = BE and BE = CF
By this we can deduce that AD//CF and AB=CF
If AD//CF and AD=CF, then the distance between these line segments will also be equal (i.e. AC = DF) and //. Therefore ACFD is a //gm
AC = DF (ACFD is a //gm as proved above)
In triangles ABC and DEF
AB = DE (given)
BC = EF (given)
AC = DF (proved above)
Triangle ABC ≅Triangle DEF by SSS Criteria
Given:
To prove:
Proof:
AD = BC (given)
∠A = ∠B (Equal sides have opposite angles equal)
In trapezium ABCD
∠A + ∠D = 1800 (AB//DC, AD is the transversal; they are interior angles) … (1)
∠B + ∠C = 1800 (AB//DC, BC is the transversal; they are interior angles) … (2)
SUM of interior angles s always supplementary
Equating (1) and (2)
∠A + ∠D = ∠B +∠ C
But ∠A = ∠B (proved above)
Therefore ∠D = ∠C
In triangles ABC and BAD
AB = AB (common)
∠A = ∠B (proved above)
AD = BC (given)
Triangle ABC ≅Triangle BAD by SAS Criteria
Diagonal AC = Diagonal BD (by CPCT)
NCERT solutions for class 9 Maths Chapter 8 Exercise 8.2
Given:
To prove:
Proof:
In triangle ADC, S and R are the mid points of AD and DC
According to mid-point theorem, the line joining the mid-points of two sides of the triangle is // to the third side and is half of it.
Therefore, SR//AC and SR = 1/2 AC … (1)
Similarly using mid-point theorem on triangle ABC, we get
PQ//AC and PQ = 1/2 AC … (2)
By (1) and (2)
SR//AC and PQ//AC, and
SR = 1/2 AC and PQ = 1/2 AC, therefore, SR//PQ and SR = PQ … (A)
(iii) Since SR//PQ and SR = PQ, that means that lines formed by joining the end points of these line segments will also be // and equal in length
Therefore, PQRS is a //gm
Alternate method for part (iii)
Construct second diagonal BD
By using midpoint theorem on triangles DBC and ABD, we get
RQ//BD and RQ= 1/2 BD … (3)
SP//BD and SP = 1/2 BD … (4)
By (3) and (4),
SP//RQ and SP=RQ … (B)
By (A) and (B), it’s clear that PQRS is a //gm
Given:
To prove:
PQRS is a rectangle:
Construct diagonal AC
In triangle ABC, P and Q are the mid points of AB and BC
According to mid-point theorem, the line joining the mid-points of two sides of the triangle is // to the third side and is half of it.
Therefore, PQ//AC and PQ = 1/2 AC … (1)
Similarly using mid-point theorem on triangle ADC, we get
RS//AC and RS = 1/2 AC … (2)
By (1) and (2) we get
SR//AC and PQ//AC
SR = 1/2 AC and PQ = 1/2 AC, therefore, SR//PQ and SR = PQ … (A)
Construct second diagonal BD
By using midpoint theorem on triangles DBC and ABD, we get
RQ//BD and RQ= 1/2 BD … (3)
SP//BD and SP = 1/2 BD … (4)
By (3) and (4),
SP//RQ and SP=RQ … (B)
By (A) and (B), it’s clear that PQRS is a //gm
Given:
To prove:
PQRS is a rhombus
Proof:
Construct diagonal AC
In triangle ABC, P and Q are the mid points of AB and BC
According to mid-point theorem, the line joining the mid-points of two sides of the triangle is // to the third side and is half of it.
Therefore, PQ//AC and PQ = 1/2 AC … (1)
Similarly using mid-point theorem on triangle ADC, we get
RS//AC and RS = 1/2 AC … (2)
By (1) and (2) we get
SR//AC and PQ//AC
SR = 1/2 AC and PQ = 1/2 AC, therefore, SR//PQ and SR = PQ=1/2 AC … (A)
Construct second diagonal BD
By using midpoint theorem on triangles DBC and ABD, we get
RQ//BD and RQ= 1/2 BD … (3)
SP//BD and SP = 1/2 BD … (4)
By (3) and (4),
SP//RQ and SP=RQ = 1/2 BD … (B)
By (A) and (B) we can state that PQRS is a //gm
But diagonals of a rectangle are equal that is, AC = BD
So, (B) can be written as SP=RQ = 1/2 AC … (C)
By (A) and (C), we can deduce that PQ =QR = RS = SP, therefore // PQRS is a rhombus.
Given:
Proof: In Triangle ADB, E is the midpoint of AD and EF //AB, therefore EO is also // to AB (as it is a part of EF), by converse of midpoint theorem, O is midpoint of DB.
Similarly in triangle BCD, O is the midpoint of BD and OF//AB, therefore by converse of midpoint theorem F is also the midpoint of BC.
Given:
To prove: AF and EC trisect diagonal BD
Proof:
In Triangle APB, as shown in figure, we can observe that EQ//AP, and E is the midpoint of AB, therefore by midpoint theorem, Q is the midpoint of BP. That is, PQ=QB … (1)
Similarly in triangle DQC, FP//QC, and F is the midpoint of DC, so by converse of midpoint theorem, P is the midpoint of DQ. That is, DP =PQ … (2)
By (1) and (2)
DP= PQ = QB, therefore, P and Q trisect diagonal BD.
M is the midpoint of AB and DM//BC, therefore by converse of midpoint theorem, D is the midpoint of AC
BC//DM, ∠ACB = ∠ADM = 900 (corresponding angles). This proves that MD is perpendicular to AC
Draw a line // to AC from A, extend the line CM to form a vertex F. It forms a rectangle.
Diagonals of a rectangle are equal and they bisect each other
AB = CF
Or AB = 2 CM => CM= 1/2 AB … (1)
Since M is midpoint of AB, MA = MB = 1/2 AB …(2)
By (1) and (2)
MA = CM= 1/2 AB
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